Javascript I'；我正试图将数据从数据库拉入选择框

我试图将数据从数据库拉入选择框，但当数据被拉入一个“td”而不是单独的td时。我正在努力实现如下所示的结果

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但我一直得到这个结果

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这是我的密码

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您需要将整个表行放入循环中。您还需要添加一个变量来计算行号

<?php 
$row =1;
foreach($data_array as $rows2):
?>
<tr>
    <td class="count"><?php echo $row; ?></td>
    <td>
        <?php    
             $fighterID = $rows2->FireFighterInfo_fighterID; 
            $results = mysql_query("SELECT `fighterID`, `firstName`, `middleName`, `lastName`, `stationlocation`.`exactlocation` 
                                    FROM `firefighterinfo` 
                                    JOIN `stationlocation` ON `firefighterinfo`.`StationLocation_locationID`=`stationlocation`.`locationID` ORDER BY `firstName`");               
            echo '<select name="fireman[]" required><option value=""></option>';
            while($row = mysql_fetch_array($results))
            {
                if($row['fighterID'] == $fighterID)
                    echo '<option selected>'. $row['firstName']." ".$row['middleName']." ".$row['lastName'].", ".$row['exactlocation'].'</option>';
                else
                    echo '<option>'. $row['firstName']." ".$row['middleName']." ".$row['lastName'].", ".$row['exactlocation'].'</option>';

            }// end while
            echo '</select><br>';

        ?> 
    </td>
    <td>
        <input type="button" value="X" class="removeVar"/>
    </td>
</tr>
<?php
    $row++;
    endforeach;
?>    


将PHP代码包含到for loop中。嗯，为什么要在JS中嵌入PHP代码？您知道Javascript在客户端执行，而PHP在服务器上执行吗？在加载页面时，JS代码将被硬编码到php代码的输出中，然后再也不会更改。这在某种程度上是可行的。我两次得到“1”，第一行没有选择框，然后当按下“添加项目”按钮时，编号跳到数字5。你建议我做什么？
<script type="text/javascript">
    $('form').on('click', '.removeVar', function(){
    $(this).closest('tr').remove();
    $('.count').each(function(i){
      $(this).text(i + 1);
    });
});

//add a new node
$('#addVar').on('click', function(){
    var varCount = $('#myTable tr').length - 1;
    $node = ['<tr>',
    '<td class="count">'+varCount+'</td>',
    '<td><select name="fireman[]" class="ctlGroup" required>',
      '<option value=""></option>',
      '<?php require("php/fireman_list.php"); ?>',
    '</select></td>',
    '<td><input type="button" value="X" class="removeVar"/>',
    '</td></tr>'].join('\n');
    $('#myTable > tbody:last').append($node);
});
</script>

<?php 
$row =1;
foreach($data_array as $rows2):
?>
<tr>
    <td class="count"><?php echo $row; ?></td>
    <td>
        <?php    
             $fighterID = $rows2->FireFighterInfo_fighterID; 
            $results = mysql_query("SELECT `fighterID`, `firstName`, `middleName`, `lastName`, `stationlocation`.`exactlocation` 
                                    FROM `firefighterinfo` 
                                    JOIN `stationlocation` ON `firefighterinfo`.`StationLocation_locationID`=`stationlocation`.`locationID` ORDER BY `firstName`");               
            echo '<select name="fireman[]" required><option value=""></option>';
            while($row = mysql_fetch_array($results))
            {
                if($row['fighterID'] == $fighterID)
                    echo '<option selected>'. $row['firstName']." ".$row['middleName']." ".$row['lastName'].", ".$row['exactlocation'].'</option>';
                else
                    echo '<option>'. $row['firstName']." ".$row['middleName']." ".$row['lastName'].", ".$row['exactlocation'].'</option>';

            }// end while
            echo '</select><br>';

        ?> 
    </td>
    <td>
        <input type="button" value="X" class="removeVar"/>
    </td>
</tr>
<?php
    $row++;
    endforeach;
?>